結果
問題 | No.872 All Tree Path |
ユーザー | Haar |
提出日時 | 2019-08-30 22:12:11 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 195 ms / 3,000 ms |
コード長 | 3,217 bytes |
コンパイル時間 | 2,198 ms |
コンパイル使用メモリ | 205,688 KB |
実行使用メモリ | 30,464 KB |
最終ジャッジ日時 | 2024-11-21 23:55:59 |
合計ジャッジ時間 | 4,753 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 188 ms
20,096 KB |
testcase_01 | AC | 184 ms
20,096 KB |
testcase_02 | AC | 192 ms
20,096 KB |
testcase_03 | AC | 98 ms
30,464 KB |
testcase_04 | AC | 2 ms
5,248 KB |
testcase_05 | AC | 195 ms
20,096 KB |
testcase_06 | AC | 194 ms
19,968 KB |
testcase_07 | AC | 176 ms
20,096 KB |
testcase_08 | AC | 13 ms
5,248 KB |
testcase_09 | AC | 13 ms
5,248 KB |
testcase_10 | AC | 12 ms
5,248 KB |
testcase_11 | AC | 13 ms
5,248 KB |
testcase_12 | AC | 12 ms
5,248 KB |
testcase_13 | AC | 2 ms
5,248 KB |
testcase_14 | AC | 2 ms
5,248 KB |
testcase_15 | AC | 2 ms
5,248 KB |
testcase_16 | AC | 2 ms
5,248 KB |
testcase_17 | AC | 2 ms
5,248 KB |
testcase_18 | AC | 2 ms
5,248 KB |
testcase_19 | AC | 2 ms
5,248 KB |
ソースコード
#include <bits/stdc++.h> #define LLI long long int #define FOR(v, a, b) for(LLI v = (a); v < (b); ++v) #define FORE(v, a, b) for(LLI v = (a); v <= (b); ++v) #define REP(v, n) FOR(v, 0, n) #define REPE(v, n) FORE(v, 0, n) #define REV(v, a, b) for(LLI v = (a); v >= (b); --v) #define ALL(x) (x).begin(), (x).end() #define RALL(x) (x).rbegin(), (x).rend() #define ITR(it, c) for(auto it = (c).begin(); it != (c).end(); ++it) #define RITR(it, c) for(auto it = (c).rbegin(); it != (c).rend(); ++it) #define EXIST(c,x) ((c).find(x) != (c).end()) #define fst first #define snd second #define popcount __builtin_popcount #define UNIQ(v) (v).erase(unique(ALL(v)), (v).end()) #define bit(i) (1LL<<(i)) #ifdef DEBUG #include <misc/C++/Debug.cpp> #else #define dump(...) ((void)0) #endif #define gcd __gcd using namespace std; template <class T> constexpr T lcm(T m, T n){return m/gcd(m,n)*n;} template <typename I> void join(ostream &ost, I s, I t, string d=" "){for(auto i=s; i!=t; ++i){if(i!=s)ost<<d; ost<<*i;}ost<<endl;} template <typename T> istream& operator>>(istream &is, vector<T> &v){for(auto &a : v) is >> a; return is;} template <typename T, typename U> bool chmin(T &a, const U &b){return (a>b ? a=b, true : false);} template <typename T, typename U> bool chmax(T &a, const U &b){return (a<b ? a=b, true : false);} template <typename T, size_t N, typename U> void fill_array(T (&a)[N], const U &v){fill((U*)a, (U*)(a+N), v);} struct Init{ Init(){ cin.tie(0); ios::sync_with_stdio(false); } }init; template <typename Cost = int> class Edge{ public: int from,to; Cost cost; Edge() {} Edge(int to, Cost cost): to(to), cost(cost){} Edge(int from, int to, Cost cost): from(from), to(to), cost(cost){} Edge rev() const {return Edge(to,from,cost);} friend ostream& operator<<(ostream &os, const Edge &e){ os << "(FROM: " << e.from << "," << "TO: " << e.to << "," << "COST: " << e.cost << ")"; return os; } }; template <typename T> using Graph = vector<vector<Edge<T>>>; template <typename T> using Tree = vector<vector<Edge<T>>>; template <typename C, typename T> void add_edge(C &g, int from, int to, T w){ g[from].push_back(Edge<T>(from, to, w)); } template <typename C, typename T> void add_undirected(C &g, int a, int b, T w){ g[a].push_back(Edge<T>(a, b, w)); g[b].push_back(Edge<T>(b, a, w)); } Tree<int> tree; int N; int sub[202000]; LLI dp1[202000]; LLI dp2[202000]; void dfs(int cur, int par){ dp1[cur] = 0; sub[cur] = 1; for(auto &e : tree[cur]){ if(e.to == par) continue; dfs(e.to, cur); sub[cur] += sub[e.to]; dp1[cur] += dp1[e.to] + sub[e.to] * e.cost; } } void dfs2(int cur, int par, LLI cost){ if(par != -1){ dp2[cur] += dp2[par]; dp2[cur] += dp1[par] - (dp1[cur] + sub[cur] * cost); } dp2[cur] += (N-sub[cur]) * cost; for(auto &e : tree[cur]){ if(e.to == par) continue; dfs2(e.to, cur, e.cost); } } int main(){ cin >> N; tree = Tree<int>(N); REP(i,N-1){ int u,v,w; cin >> u >> v >> w; --u, --v; add_undirected(tree, u, v, w); } dfs(0,-1); dfs2(0,-1,0); LLI ans = 0; REP(i,N) ans += dp1[i] + dp2[i]; cout << ans << endl; return 0; }